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Chapter 5: Problem 576

In a bicycle the radius of rear wheel is twice the radius of front wheel. If \(\mathrm{r}_{\mathrm{F}}\) and \(\mathrm{r}_{\mathrm{r}}\) are the radius, \(\mathrm{v}_{\mathrm{F}}\) and \(\mathrm{v}_{\mathrm{r}}\) are speed of top most points of wheel respectively then... \(\\{\mathrm{A}\\} \mathrm{v}_{\mathrm{r}}=2 \mathrm{v}_{\mathrm{F}}\) $\\{\mathrm{B}\\} \mathrm{v}_{\mathrm{F}}=2 \mathrm{v}_{\mathrm{r}} \quad\\{\mathrm{C}\\} \mathrm{v}_{\mathrm{F}}=\mathrm{v}_{\mathrm{r}}$ \(\\{\mathrm{D}\\} \mathrm{v}_{\mathrm{F}}>\mathrm{v}_{\mathrm{r}}\)

Short Answer

Expert verified
The correct answer is option A: \(v_r = 2v_F\).

Step by step solution

01

Write down the given information

We are given the following: 1. The radius of the rear wheel is twice the radius of the front wheel: \(r_r = 2r_F\) 2. The speeds of the top points of the wheels are \(v_F\) and \(v_r\).
02

Find the relationship between the linear and angular velocities of the two wheels

To find the relationship between the wheels' speeds, we will use the formula connecting linear and angular velocity: \(v = r\omega\), where \(v\) is the linear velocity, \(r\) is the radius, and \(\omega\) is the angular velocity. For the front wheel, we have: \[v_F = r_F \omega_F\] For the rear wheel, we have: \[v_r = r_r \omega_r\]
03

Use the given information to relate the angular velocities

Since the radius of the rear wheel is twice the radius of the front wheel, we can create an expression for the angular velocities in terms of linear velocities: \[\omega_F = \frac{v_F}{r_F}\] and \[\omega_r = \frac{v_r}{r_r} = \frac{v_r}{2r_F}\] Now, note that both the wheels are attached to the same bicycle, which means they will always be in contact with the ground. Hence, their angular velocities will be equal: \[\omega_F = \omega_r\]
04

Solve for the relationship between the linear velocities

Substitute the expressions for \(\omega_F\) and \(\omega_r\) from Step 3 into the equation \(\omega_F = \omega_r\): \[\frac{v_F}{r_F} = \frac{v_r}{2r_F}\] Now, solve for the relationship between \(v_F\) and \(v_r\): \(v_F = \frac{1}{2}v_r\) Which is equivalent to: \(v_r = 2v_F\)
05

Identify the correct answer

Comparing our result with the given options, we find that our result matches option A: \(v_r = 2v_F\) Hence, the correct answer is option A.

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Most popular questions from this chapter

A thin circular ring of mass \(\mathrm{M}\) and radius \(\mathrm{r}\) is rotating about its axis with a constant angular velocity \(\mathrm{w}\). Two objects each of mass \(\mathrm{m}\) are attached gently to the opposite ends of a diameter of the ring. The ring will now rotate with an angular velocity.... $\\{\mathrm{A}\\}[\\{\omega(\mathrm{M}-2 \mathrm{~m})\\} /\\{\mathrm{M}+2 \mathrm{~m}\\}]$ \(\\{\mathrm{B}\\}[\\{\omega \mathrm{M}\\} /\\{\mathrm{M}+2 \mathrm{~m}\\}]\) \(\\{C\\}[\\{\omega M)\\} /\\{M+m\\}]\) \(\\{\mathrm{D}\\}[\\{\omega(\mathrm{M}+2 \mathrm{~m})\\} / \mathrm{M}]\)

A circular disc of radius \(\mathrm{R}\) and thickness \(\mathrm{R} / 6\) has moment of inertia I about an axis passing through its centre and perpendicular to its plane. It is melted and re-casted in to a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation is \(\ldots\) \(\\{\mathrm{A}\\} \mathrm{I}\) \(\\{\mathrm{B}\\}(2 \mathrm{I} / 8)\) \(\\{\mathrm{C}\\}(\mathrm{I} / 5)\) \(\\{\mathrm{D}\\}(\mathrm{I} / 10)\)

Two identical hollow spheres of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) are joined together and the combination is rotated about an axis tangential to one sphere and perpendicular to the line connecting their centres. The moment of inertia of the combination is \(\\{\mathrm{A}\\} 10 \mathrm{MR}^{2}\) \(\\{\mathrm{B}\\}(4 / 3) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(32 / 3) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(34 / 3) \mathrm{MR}^{2}\)

Match list I with list II and select the correct answer $$ \begin{aligned} &\begin{array}{|l|l|} \hline \text { List-I } & \begin{array}{l} \text { List - II } \\ \text { System } \end{array} & \text { Moment of inertia } \\ \hline \text { (x) A ring about it axis } & \text { (1) }\left(\mathrm{MR}^{2} / 2\right) \\ \hline \text { (y) A uniform circular disc about it axis } & \text { (2) }(2 / 5) \mathrm{MR}^{2} \\ \hline \text { (z) A solid sphere about any diameter } & \text { (3) }(7 / 5) \mathrm{MR}^{2} \\ \hline \text { (w) A solid sphere about any tangent } & \text { (4) } \mathrm{MR}^{2} \\ \cline { 2 } & \text { (5) }(9 / 5) \mathrm{MR}^{2} \\ \hline \end{array}\\\ &\text { Select correct option }\\\ &\begin{array}{|l|l|l|l|l|} \hline \text { Option? } & \mathrm{X} & \mathrm{Y} & \mathrm{Z} & \mathrm{W} \\\ \hline\\{\mathrm{A}\\} & 2 & 1 & 3 & 4 \\ \hline\\{\mathrm{B}\\} & 4 & 3 & 2 & 5 \\ \hline\\{\mathrm{C}\\} & 1 & 5 & 4 & 3 \\ \hline\\{\mathrm{D}\\} & 4 & 1 & 2 & 3 \\ \hline \end{array} \end{aligned} $$

The moment of inertia of a uniform circular disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) about any of its diameter is \((1 / 4) \mathrm{MR}^{2}\), what is the moment of inertia of the disc about an axis passing through its centre and normal to the disc? \(\\{\mathrm{A}\\} \mathrm{MR}^{2}\) \\{B \(\\}(1 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(3 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\} 2 \mathrm{MR}^{2}\)
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